# Study Notes: Physics Review Problems --- ## Problem 1: Planet Orbiting a Star (Elliptical Orbit) ### Key Concepts: - **Elliptical Orbits**: A planet's orbit around a star is elliptical, with the closest point to the star called **periapsis (A)** and the farthest point called **apoapsis (B)**. - **Moment of Inertia ($I$)**: For a planet of mass $m$, $I = mr^2$, where $r$ is the distance between the planet and the star. --- ### (a) What does $r$ represent in this situation? $r$ represents the distance between the center of the planet and the center of the star. --- ### (b) Why is the linear speed greater at A than at B? Using **Conservation of Angular Momentum**: 1. Angular momentum $L$ is conserved: $$L = I\omega = mr^2v$$ where $v$ is the linear speed and $\omega$ is the angular velocity. 2. At points A and B: $$L_A = L_B \implies m r_A^2 v_A = m r_B^2 v_B$$ 3. Simplify: $$r_A v_A = r_B v_B$$ Since $r_A < r_B$, it follows that $v_A > v_B$. Angular momentum is conserved because gravity acts as a central force, producing no net torque. --- ### (c) Why is linear momentum not conserved, but angular momentum is? - **Linear Momentum**: Not conserved because gravity is an external force acting on the planet, changing its linear speed. - **Angular Momentum**: Conserved because gravity acts radially (towards the center), producing no torque. Students often assume linear momentum is conserved in orbital motion. Remember, external forces like gravity can change linear momentum. --- ## Problem 2: Rotating Disk with Torque ### Key Concepts: - **Moment of Inertia of a Disk**: $$I = \frac{1}{2}MR^2$$ where $M$ is the mass and $R$ is the radius of the disk. - **Torque ($\tau$)**: $$\tau = I\alpha$$ where $\alpha$ is the angular acceleration. --- ### (a) Angular Acceleration of the Disk From the graph, angular speed $\omega$ increases linearly with time $t$. Angular acceleration $\alpha$ is the slope of the $\omega$ vs. $t$ graph: $$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{12.5}{2.0} = 6.25 \, \text{rad/s}^2$$ --- ### (b) Torque Acting on the Disk Using $\tau = I\alpha$: 1. Moment of inertia: $$I = \frac{1}{2}MR^2 = \frac{1}{2}(0.20)(0.40)^2 = 0.016 \, \text{kg·m}^2$$ 2. Torque: $$\tau = I\alpha = (0.016)(6.25) = 0.10 \, \text{Nm}$$ --- ### (c.i) New Angular Speed After Object Falls on Disk 1. Initial angular momentum: $$L_{\text{initial}} = I_{\text{disk}} \omega = (0.016)(12.5) = 0.20 \, \text{kg·m}^2/\text{s}$$ 2. Moment of inertia after object sticks: $$I_{\text{new}} = I_{\text{disk}} + mR^2 = 0.016 + (0.10)(0.40)^2 = 0.032 \, \text{kg·m}^2$$ 3. Conservation of angular momentum: $$L_{\text{initial}} = L_{\text{final}} \implies I_{\text{new}} \omega' = 0.20$$ $$\omega' = \frac{0.20}{0.032} = 6.25 \, \text{rad/s}$$ --- ### (c.ii) Fraction of Energy Lost 1. Initial kinetic energy: $$KE_{\text{initial}} = \frac{1}{2}I_{\text{disk}}\omega^2 = \frac{1}{2}(0.016)(12.5)^2 = 1.25 \, \text{J}$$ 2. Final kinetic energy: $$KE_{\text{final}} = \frac{1}{2}I_{\text{new}}\omega'^2 = \frac{1}{2}(0.032)(6.25)^2 = 0.625 \, \text{J}$$ 3. Fraction lost: $$\text{Fraction lost} = \frac{KE_{\text{initial}} - KE_{\text{final}}}{KE_{\text{initial}}} = \frac{1.25 - 0.625}{1.25} = 0.50 \, (50\%)$$ Energy is lost due to the inelastic collision when the object sticks to the disk. --- ## Problem 3: Spaceship Traveling at Relativistic Speeds ### Key Concepts: - **Length Contraction**: $$L = L_0 \sqrt{1 - v^2/c^2}$$ where $L_0$ is the proper length, $v$ is the velocity, and $c$ is the speed of light. - **Time Dilation**: $$\Delta t = \gamma \Delta t_0 \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$ --- ### (a) Distance Traveled in Spaceship Frame 1. Lorentz factor: $$\gamma = \frac{1}{\sqrt{1 - (0.4)^2}} = 1.09$$ 2. Contracted distance: $$L = \frac{L_0}{\gamma} = \frac{2.0 \times 10^8}{1.09} = 1.8 \times 10^8 \, \text{km}$$ --- ### (b.i) Time for Signal to Reach Spaceship (Earth Frame) 1. Distance to spaceship: $d = 3.0 \times 10^8 \, \text{km}$ 2. Relative velocity of signal: $c - v = 3.0 \times 10^8 - 0.4c = 0.6c$ 3. Time: $$\Delta t = \frac{d}{c - v} = \frac{3.0 \times 10^8}{0.6c} = 17 \, \text{s}$$ --- ### (b.ii) Time for Signal in Spaceship Frame 1. Time dilation: $$\Delta t' = \gamma (\Delta t - \frac{vL}{c^2})$$ Substituting values: $$\Delta t' = 1.09 \left(16.7 - \frac{0.4c \cdot 3.0 \times 10^8}{c^2}\right) = 11 \, \text{s}$$ Always account for relativistic effects like time dilation and length contraction when dealing with high velocities. --- ## Problem 4: Spacecraft Traveling to a Planet ### Key Concepts: - **Relativistic Velocity Addition**: $$v_{\text{rel}} = \frac{v_1 + v_2}{1 + \frac{v_1v_2}{c^2}}$$ --- ### (a.i) Time of Arrival (Earth Frame) 1. Distance: $12.0 \, \text{ly}$ 2. Speed: $0.8c$ 3. Time: $$t = \frac{d}{v} = \frac{12.0}{0.8} = 15.0 \, \text{years}$$ --- ### (a.ii) Time of Arrival (Spacecraft Frame) 1. Lorentz factor: $$\gamma = \frac{1}{\sqrt{1 - (0.8)^2}} = 1.67$$ 2. Time: $$t' = \frac{t}{\gamma} = \frac{15.0}{1.67} = 9.0 \, \text{years}$$ --- ### (b.ii) Relative Speed of Cargo Ship and Spacecraft Using relativistic velocity addition: $$v_{\text{rel}} = \frac{v_{\text{cargo}} - v_{\text{spacecraft}}}{1 - \frac{v_{\text{cargo}}v_{\text{spacecraft}}}{c^2}}$$ Substitute: $$v_{\text{rel}} = \frac{0.6c - (-0.8c)}{1 - \frac{(0.6c)(-0.8c)}{c^2}} = 0.946c$$ --- ### (b.iii) Length of Cargo Ship in Spacecraft Frame 1. Lorentz factor: $$\gamma = \frac{1}{\sqrt{1 - (0.946)^2}} = 3.08$$ 2. Contracted length: $$L = \frac{L_0}{\gamma} = \frac{992}{3.08} = 322 \, \text{m}$$ Proper length is the length measured in the rest frame of the object. --- ### Key Takeaways: - Angular momentum is conserved in central force systems, but linear momentum is not. - Torque and angular acceleration are directly related through the moment of inertia. - Relativistic effects like time dilation and length contraction are crucial at high velocities. - Energy is not always conserved in inelastic collisions, but angular momentum is.